# Everything Said Twice > Every word leaves a mark. Count who keeps coming back. Canonical URL: Domain: Python · Difficulty: easy · Seniority: L3 ## Problem Given a string of words separated by whitespace, return a dict mapping each distinct word to the number of times it appears. ## Worked solution and explanation ### What this is really testing Strip the costume and this is a tokenize-and-tally: chop a string into words, count how many times each shows up. Everyone gets the counting part right; almost nobody trips on the dict accumulation. The thing that actually separates people is one keystroke: whether you call text.split() with no argument or text.split(' ') with a literal space. They look identical on the happy path 'a b c' and diverge the instant the input has a double space, a tab, or a leading blank. Pick the wrong one and you silently seed your dict with an empty-string key that carries a real count, and your output is wrong in a way that passes casual eyeballing. > **The one insight** > > str.split() with no argument is a completely different function from str.split(' '). No-arg split treats any run of whitespace as a single delimiter AND trims the ends, so ' the the cat '.split() is exactly ['the', 'the', 'cat']. That single behavior handles every whitespace trap in the problem for free. Reach for it and the rest is a one-liner. ### How you get there #### Step 1: Split on whitespace, not on a space character The spec says 'separated by whitespace' and that phrasing is the tell. No-arg split() collapses runs of spaces, tabs, and newlines, and strips leading/trailing whitespace, so you never manufacture empty tokens. The moment you hardcode split(' ') you are counting on the input being perfectly single-spaced, which real text never is. #### Step 2: Tally in one pass Feed the token list straight into collections.Counter. It walks the tokens once, O(n), and increments in C. You could hand-roll d[w] = d.get(w, 0) + 1 and it would be correct, but Counter is the idiom a reviewer expects here and it comes with .most_common() for the inevitable follow-up. #### Step 3: Hand back a plain dict The signature promises -> dict. Counter IS a dict subclass and compares equal, so tests usually pass either way, but wrapping in dict() matches the contract exactly and won't surprise a caller that inspects the concrete type. Empty input needs no special case: ''.split() is [], Counter([]) is empty, dict() is {}. **Counter over whitespace-split tokens** ```python from collections import Counter def word_counts(text: str) -> dict: return dict(Counter(text.split())) ``` *One pass to split, one pass to count.* **split(' ') — the trap** ' the the cat '.split(' ') -> ['', '', 'the', '', 'the', 'cat', ''] . Counter now has a '' key with count 3 and 'the' count 2. The output silently carries a junk key and passes a quick glance. **split() — correct** ' the the cat '.split() -> ['the', 'the', 'cat'] . Counter is {'the': 2, 'cat': 1} . Runs collapse, ends trim, no empty tokens, ever. > **Common pitfall** > > Two ways candidates lose this. First: split(' ') with an explicit space, which injects empty-string keys on any padded or multi-spaced input like ' click view '. Second: assuming case folds — 'Go go GO' is three distinct words here, not one. Do NOT sneak in a .lower() unless the spec asks; the hidden tests expect GO, Go, and go as separate keys. > **Interviewers watch for** > > The tell of seniority is reaching for Counter without hesitation, split() with no arg, and naming the empty-input case out loud before being asked. Bonus signal: mentioning .most_common(k) exists the second the interviewer starts to say 'now give me the top…'. > **Why it stays cheap** > > O(n) in the length of text: split scans the string once, Counter scans the tokens once. Space is O(u) for u distinct words. Counter's increment loop runs in C, so it edges out a Python-level d.get() loop by a comfortable margin — but the real win is that there's no second pass and no sort unless a follow-up demands ranking. > **In production** > > This exact shape is everywhere: event-tag frequency, log-level counts, word clouds, n-gram tallies. The whitespace trap is the same one that corrupts real analytics when someone splits log lines on ' ' and a tab-delimited field quietly produces phantom empty tokens in the histogram. ## Common follow-up questions - What changes if words should be counted case-insensitively? _(Counter(text.lower().split()) — but note .lower() allocates a full copy, O(n) extra memory, and check whether the spec actually wants case folded.)_ - How would you return the top 3 most common words? _(Counter(...).most_common(3) gives (word, count) tuples in descending order; be ready to discuss that ties break on insertion order.)_ - How would you handle a 10 GB file that doesn't fit in memory? _(Stream line by line, split each line, Counter.update(tokens) in place; the Counter stays bounded by the vocabulary size, not the file size.)_ ## Related - [All practice problems](https://datadriven.io/problems) - [Mock interview mode](https://datadriven.io/interview/everything_said_twice) - [Python Interview Questions](https://datadriven.io/python-interview-questions) - [Data Engineering Interview Prep Guide](https://datadriven.io/data-engineer-interview-prep) - [Daily Challenge](https://datadriven.io/daily) --- Source: DataDriven (https://datadriven.io). DataDriven is the data engineering interview community. Live code execution in SQL, Python, and Spark sandboxes. Every feature is open to every member.